UNIT 1: ELECTROSTATICS

Transcription

1 SYLLABUS PART- A UNIT 1: ELECTROSTATICS Experimental law of coulomb, Electric field intensity, Field due to continuous volume charge distribution, Field of a line charge, Electric flux density, Electric flux density, Gauss law, Divergence, Vector Operator, Divergence theorem 6 Hours UNIT : ENERGY & POTENTIAL, CONDUCTORS, DIELECTRICS & CAPACITANCE (a).energy expended in moving a point charge in an electric field, Line integral, Definition of potential difference and potential, Potential field of a point charge & system of charges, Potential gradient, Energy density in an electrostatic field. (b).current and current density, Continuity of current, metallic conductors, Conductor properties and boundary conditions for dielectrics, Conductor properties and boundary conditions for perfect dielectrics, capacitance and examples 8 Hours UNIT 3: POISSONS AND LAPLACES EQUATIONS Derivation of Poisson s equation and Laplace s equation, Uniqueness theorem, Examples of the solutions Laplace Equations and Poisson s Equations 6 Hours UNIT 4: THE STEADY MAGNETIC FIELDS Biot- savart law, Amphere s circuitary law, curl, Stoke s theorem, Magnetic flux and flux density, Scalar and Vector magnetic potentials. 6 Hours PART B UNIT 5 : MAGNETIC FORCES, MATERIALS AND INDUCTANCE (a).force on a moving charge and differential current element, Force between differential current elements, Force and Torque on a closed circuit, (b).magnetic materials and inductance, Magnetization and permeability, Magnetic boundary conditions, Magnetic circuits, Potential energy and forces on magnetic materials, Inductance and mutual inductance 8 Hours Dept of E&C, SJBIT 1

2 UNIT - 6: TIME VARYING MAGENTIC FIELDS AND MAXWELLS EQUATION Faraday s law, Displacement current, Maxwell s equation in point and integral form, Retarded potentials. 6 Hours UNIT - 7: UNIFORM PLANE WAVES Wave propagation in free space and dielectrics, Poynting s theorem and wave power, Propagation in good conductors (skin effect) 6 Hours UNIT - 8: PLANE WAVES AT BOUNDARIES AND DISPERSIVE MEDIA Reflection of uniform plane waves at normal incidence, SWR, Plane wave propagation in general directions 6 Hours TEXT BOOK: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006. REFERENCE BOOKS:. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Dept of E&C, SJBIT

3 INDEX SHEET SL.NO TOPIC PAGE NO. PART A UNIT 1: ELECTROSTATICS Experimental Law of Coulomb 5 Force on a point charge 6 3 Force due to several charges 6 4 Electric field intensity 7 5 Electric Field intensity due to several charges 7 6 Electric Field intensity at a point due to infinite sheet of charge 7 7 Electric Field at a point on the axis at a charges circular ring 9 8 Electric Flux 10 9 Electric Flux Density Gauss s Law 1 11 Application of Gauss Law 14 1 Divergence and Maxwellls First Equation 16 UNIT : ENERGY AND POTENTIAL, CONDUCTORS, 0-31 DIELECTRICS AND CAPACITANCE 13 Energy expended in moving a point charge in an electric field 0 14 Definition of Potential and Potential difference 1 15 Equipotential surface, Potential field of a point charge 16 Potential field of system of charges 3 17 Potential gradient 4 18 Energy density in an electrostatic field 7 19 Potential energy in a continuous charge distribution 8 0 Boundary condition for conductor free space interface 8 1 Boundary condition for perfect dielectric 31 UNIT 3: POISSON S AND LAPLACES EQUATIONS Laplace s and Poisson s Equation 34 3 Uniqueness Theorem 35 4 Examples of solution of Laplace s equation 36 UNIT 4 :THE STEADY MAGNETIC FIELD Introduction 40 6 Biot- Savart s Law 41 7 Amperes Circuital Law 47 8 Application of Amperes law 48 9 Magnetic Flux Density Magnetic scalar & vector potentials 50 Dept of E&C, SJBIT 3

4 PART B UNIT 5 : MAGNETIC FORCES, MATERIALS AND INDUCTANCE 31 Force on a charged particle 54 3 Force between two current elements Magnetic Torque and Moment Stokes Theorem Force on a moving charge due to electric and magnetic fields Force on a current element in a magnetic field Boundary conditions on H and B 6 38 Scalar magnetic potential Vector Magnetic potential Force and Torque on a loop or coil Materials in magnetic fields 70 4 Inductance 73 UNIT - 6 :TIME VARYING FIELDS AND MAXWELLS EQUATION Introduction Transformer and motional emf Displacement Current Equation of continuity for time varying fields Maxwells equations for time varying fields Maxwells equations for static fields 86 UNIT 7 :UNIFORM PLANE WAVE Sinusoidal Time variation 9 50 Wave propagation in a conducting medium Wave propagation in a lossless medium Wave propagation in Dielectrics Polarization 113 UNIT 8 :PLANE WAVES AT BOUNDARIES AND IN DISPERSIVE MEDIA Reflection of uniform plane waves at normal incidence Reflection coefficient and transmission coefficient in case of normal incidence 56 Standing Wave Ratio Dept of E&C, SJBIT 4

5 UNIT 1 ELECTROSTATICS Syllabus: Experimental law of coulomb, Electric field intensity, Field due to continuous volume charge distribution, Field of a line charge, Electric flux density, Electric flux density, Gauss law, Divergence, Vector Operator, Divergence theorem REFERENCE BOOKS: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006.. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Coulomb s Law Coulomb s law states that the electrostatic force F between two point charges q1 and q is directly proportional to the product of the magnitude of the charges, and inversely proportional to the square of the distance between them., and it acts along the line joining the two charges. Then, as per the Coulomb s Law, Or F kq1q F = (kq1q)/(r²) N Where k is the constant of proportionality whose value varies with the system of units. R^ is the unit vector along the line joining the two charges. In SI unit, k=. Where is called the permittivity of the free space. It has an assigned value given as =8.834 F/m. Dept of E&C, SJBIT 5

6 Force on a point charge: The forces of attraction/repulsion between two point charges and (charges whose size is much smaller than the distance between them) are given by Coulomb s law: where m/f in SI units, and R is the distance between the two charges. Here, is the force exerted on, and is the force acting on. The unit vector points from charge toward charge 1. Accordingly,. Force on Q1 is given by F1 = Newtons F q1 q q1 q F Force due to several charges Let there be many point charges q1,q,q3...qn at distances r1,r,r3...rn from charge q. The force F1, F, F3...Fn at the charges q1,q,q3...qn respectively are: q{ rˆ + } F=Fq1+Fq+Fq3... Hence, F= q{ } N Dept of E&C, SJBIT 6

7 Electric field intensity Electric field intensity at any point in an electric field is the force experienced by positive unit charge placed at that point. Consider a charge Q located at a point A. At the point B in the electric fields set up by Q, it is required To find the electric field intensity E. Let the charge at B be and let the charge Q be fixed at A. Let r be the distance between A and B. As per the Coulomb s Law, the force between Q and q is given by: F= rˆ N If it is a unit positive charge, then by definition, F in the above equation gives the magnitude of the electric field intensity E. i.e. E=F when Therefore, the magnitude of the electric field strength is: E=Q/(4r Let r be the unit vector along the line joining A and B. Thus, the vector relation between E is written as: E=Q/(4 or²) V/m Electric Field intensity due to several charges Let there be many point charges q1,q,q3...qn at distances r1,r,r3...rn be the corresponding unit vectors. The field E1, E, E3...En at the charges q1,q,q3...qn respectively are: rˆ + E=Eq1+Eq+Eq3... Hence, E= Electric field intensity at a point due to a infinite sheet of charge Let us assume a straight line charge extending along Z axis in a cylindrical coordinate system from - to + as shown in the figure 1.1. Consider an incremental length dl at a point on the conductor. The incremental length has an incremental charge of dq= ρl dl= ρldz Coulombs. Considering the charge dq, the incremental field intensity at point p is given by, Dept of ECE, SJBIT 7

8 Where, and Therefore, Integrating the above and substituting z =ρ cot θ, we get and Dept of ECE, SJBIT 8

9 Electric field intensity at a point due to a infinite sheet of charge: Let us assume a infinite sheet of charge with surface charge density ρs as shown in the figure 1.. Divide the sheet of charge into differential width strips. number of str Consider an incremental length dl at a point on the conductor. The line charge density ρl= ρs dy. The differential Electric field intensity at point P, adding the effects of all the strips, Therefore, Electric field at a point on the axis of charged circular ring: Let ρ be the charge density of the ring. So, ρ=dq/dl dq=ρdl Electric field due to an infinitely small element = de = dq/4πεo r² rˆ Dept of ECE, SJBIT 9

10 where rˆ is the unit vector along AP. de can resolved into two rectangular components, dex and dey. Now, dex=decosθ. Taking the magnitude of de from above, the equation becomes, dex= cosθ= substituting for dq from above, we have; dex= The component dey is directed downwards. If we consider an element of the ring at a point diametrically opposite to A, then its dey component points upwards and hence, cancels with that due to element A. The dex components add up. dey=0. The total field at P is the sum of the fields due to all the elements of the ring. Therefore, E= de= dex+ dey= dex E= dex= = But, r=(r²+x²)½ Therefore, E= a x Where, a x is the unit vector along the x axis. Electric flux: The concept of electric flux is useful in association with Gauss' law. The electric flux through a planar area is defined as the electric field times the component of the area perpendicular to the field. If the area is not planar, then the evaluation of the flux generally requires an area integral since the angle will be continually changing. When the area A is used in a vector operation like this, it is understood that the magnitude of the vector is equal to the area and the direction of the vector is perpendicular to the area. Dept of ECE, SJBIT 10

11 Consider a concentric sphere having radius of a m charged up to +Q C. This sphere is then placed in another sphere having a radius of b m as shown in the figure 1.4. There is no electrical connection between them. The outer sphere is momentarily charged, then it found that the charge on the outer sphere is equal to the charge on the inner sphere. This is depicted by the radial lines. This is referred as displacement flux. Therefore, Ψ = Q. Electric flux density: If +Q C of charge on the inner sphere produces the electric flux of ψ, tthen electric flux ψ uniformly distributed over the surface area 4Πa m, where a is the radius of the inner sphere. The electric flux density si given by Similarly for the outer sphere, If the inner sphere becomes smaller and smaller retaining a charge of Q C, it becomes a point charge. The flux density at appoint r from the point charge is given by, Dept of ECE, SJBIT 11

12 The electric field intensity due to point charge in free space is given by, Therefore in free space, Gauss law: The Gauss's law states that. "The electric flux passing through any closed surface is equal to the total charge enclosed by the surface" For the Gaussian-surface shown in the following figure, the Gauss' law can be expressed mathematically,. Where Ψ = flux passing through the closed surface s =1 surface integral Ds =, flux density (vector quantity) normal to the surface Q = Total charge enclosed in the surface Gauss law for charge Q enclosed in a closed surface: Let Q be the point charge placed at the origin of imaginary sphere in spherical co-ordinate system with a radius of "a" as illustrated in the figure Dept of ECE, SJBIT 1

13 The electrical field intensity cf the point charge is found to be equal to Where r = Cl and we also know that the relation between E and D as, Therefore from (1) and () we get. at the surface of the sphere, The differential element of area on a spherical surface is, in spherical coordinate form is given by, Then the required integrand Dept of ECE, SJBIT 13

14 Then the integration over the surface as required for Gauss' law. The limits placed for integral indicate that the integration over the entire sphere in spherical coordinate system on integration we get Thus we get, comparing LHS of Gauss' law as, This indicates that, Q coulombs of electric flux are crossing the surface as the enclosed charge is Q coulombs. Application of Gauss law: In case of asymmetry, we need to choose a very closed surface such that D is almost constant over the surface. Consider any point P shown in the figure 1.6 located in the rectangular coordinate system. Dept of ECE, SJBIT 14

15 The value of D at point P, may be expressed in rectangular components as, D=D x0 a x +D y0 a y +D z0 a z.. From Gauss law, we have In order to evaluate the integral over the closed surface, the integral must be broken into six integrals, one over each surface, = +. The surface element is very small & hence D is essentially constant, Similarly,, and, Therefore collectively,. Dept of ECE, SJBIT 15

16 Charge enclosed in volume v, Divergence: From Gauss law, we know that, And applying limits, The last term in the equation is the volume charge density, ρv. We shall write it as two separate equations, And Divergence is defined as,. Statement: The flux crossing the closed surface is equal to the integral of the divergence of the flux density throughout the enclosed volume, as the volume shrinks to zero.. Dept of ECE, SJBIT 16

17 Divergence in Cartesian system, Divergence in Cylindrical system, Divergence in Spherical system, Maxwell s First equation: From divergence theorem, we have From Gauss law, Per unit volume, As the volume shrinks to zero, Therefore, div D = ρ v. Dept of ECE, SJBIT 17

18 Divergence theorem: The del operator is defined as a vector operator.. In Cartesian coordinate system, Which is equal to,. Therefore,. From Gauss law, we have And by letting, Hence we have, &. Dept of ECE, SJBIT 18

19 RECOMMENDED QUESTIONS 1. State Coulomb s law of force between any point charges & indicate the units of the quantities involved.. Derive the general expression for electric field vector due to infinite line charge using Gauss law. 3. State and prove Gauss law. 4. Derive the general expression for E at a height h(h<a), along the axis of the ring charge & normal to its plane. 5. From gauss law show that.d=σv 6. State and prove divergence theorem for symmetric condition. 7. State and prove divergence theorem for asymmetric condition. Dept of ECE, SJBIT 19

20 Syllabus: UNIT ENERGY & POTENTIAL, CONDUCTORS, DIELECTRICS AND CAPACITANCE (a).energy expended in moving a point charge in an electric field, Line integral, Definition of potential difference and potential, Potential field of a point charge & system of charges, Potential gradient, Energy density in an electrostatic field. (b).current and current density, Continuity of current, metallic conductors, Conductor properties and boundary conditions for dielectrics, Conductor properties and boundary conditions for perfect dielectrics, capacitance and examples REFERENCE BOOKS: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006.. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Energy expended in moving a point charge in an electric field Electric field intensity is defined as the force experienced by unit test charge at a point p. If the test charge is moved against the electric field, then we have to exert a force equal and opposite to that exerted by the field and this requires work to be done. Suppose we need to move a charge fo Q C a distance dl in an electric field E. The force on Q arising from the electric field is, The differential amount of work done in moving charge Q over a distance dl is given by,, as F =QE Dept of ECE, SJBIT 0

21 Thus the work done to move the charge for the finite distance is given by, Definition of Potential &Potential difference : Potential difference(v) is defined as the work done in moving unit positive charge from one point to another point in an electric field. We know that,, Therefore V=W/Q= V AB signifies potential difference between points A & B and the work done in moving the unit charge from B to A. Thus B is the initial point & A is the final point. From the previous example, the work done in moving charge Q from ρ= b to ρ= a was,.. Thus the potential difference between the points a & b is given by, Absolute electric potential is defined as the work done in moving a unit positive charge from infinity to that point against the field. Electric field is defined as force on unit charge. E= F/Q. By moving the charge Q aganist an electric field between the two points a & b work is done. Thus, Edl= Fxdl/Q =work/ charge. This work done per charge is the electric potential difference. Potential difference between points a and b at a radial distance of ra and rb from a point charge Q is given by, Dept of ECE, SJBIT 1

22 If the potential at point a is VA and at point B is VB, then. Equipotential Surface It is defined as "It is a surface composed of all- points having the same value of potential" on such surfaces no work is involved in moving a unit charge, hence no potential difference between any two points on this surface. Potential field of a point charge Consider a point charge Q to be placed in the origin of a spherical coordinate system. Consider points A & B as shown in the figure. Electric Potential difference between A & B, VAB is given by, dl in spherical co ordinate system is given the figure above and E=Q/ 4Π r. Therefore, And Dept of ECE, SJBIT

23 Potential field of system of charges: Conservative property Potential at a point has been defined as the work done in moving unit positive charge from zero reference to the point. Potential is independent of the path taken from one point to the other. Potential due to a single charge is given by V(r)= Q1/ 4Π R. If Q 1 is at r 1 & point p at r, then Potential arising from charges, Q 1 at r 1 and Q at r, is given by Potential due to n number of charges, is given by Or If point charge is a small element in the continuous volume charge distribution then, As number of point charges in the volume charge distribution tends to infinity, Similarly if the point charges takes the form of a straight line then, Similarly if the point charges takes the form of a surface charge then, Potential is a function of inverse distance. Hence we can conclude that for a zero reference at infinity, then: I Potential due to a single point charge is the work done in moving unit positive charge from zero reference to the point. Potential is independent of the path taken from one point to the other II Potential field due to number of charges is the sum of the individual potential fields arising from each charge. Dept of ECE, SJBIT 3

24 III. Potential due to continuous charge distribution is found by carrying a unit charge from infinity to the point under consideration. is independent on the path chosen for the line integral, regardless of the source of the E field. Hence we can conclude that no work is done in carrying a unit positive charge around any closed path, or Any field that satisfies an equation of the form above is said to be conservative field Potential gradient Potential at any point is given by Potential difference between points separated by a very short length L along which E is essentially constant, is given by In rectangular co ordinate system, As V is a unique function of x,y,z. Then,. Since both the expressions are true with respect dx,dy & dz, we can write, Therefore, Dept of ECE, SJBIT 4

25 In rectangular co ordinate system, Combining all the above equations allows us to use a compact expression that relates E & V, Gradient in other coordinate system is as given below, Dept of ECE, SJBIT 5

26 Dipole Dipole is the name given to point charges of equal magnitude but opposite in sign, separated by a small distance which is small compared to the distance to point P at which we need to find electric and potential fields. Dipole is as shown in the figure below, Let the distance from Q & -Q to P be R 1 & R respectively, and the distance between the two charges be d. Total Potential at point P due to Q1 & Q is given by, Since mid-way between the point charges are located at Z=0, R 1 = R & therefore the potential is zero. For distant point, R 1 R =r and R 1 & R are the approximately parallel as shown in the figure below From the figure, R -R 1 =d cos θ; Therefore, Dept of ECE, SJBIT 6

27 The plane Z=0 is at zero potential. Using gradient relationship in spherical co ordinates, From the above we obtain, Or Energy density in an electrostatic field: Consider a surface without charge. Bringing a charge Q1 from infinity to any point on the surface requires no work as there is no field present. The positioning of Q at appoint in the field of Q1 requires an amount of work to be done which is given by. Similarly work required to position each additional charge in the field is given by, Total positioning work = Potential energy of the field Bringing the charges in the reverse order, the work done is given by, Adding the energy expressions, we get For n number of charges, Dept of ECE, SJBIT 7

28 Potential energy in a continuous charge distribution: For the region with continuous charge distribution, the equation for W E= By vector identity which is true for any scalar function V & vector D, Then,, From Guass law, We can write. and from gradient Boundary condition for conductor free space interface: Consider a closed path at the boundary between conductor and a dielectric, such that h 0. We know that work done in moving a charge over a closed path is zero i.e., Therefore the integral can be broken up as,. Dept of ECE, SJBIT 8

29 . Let the length from a to b or c to d be W and from a to d or b to c be h, hence we obtain,. Hence we obtain E W=0 & therefore E t =0 Hence at the conductor dielectric interface tangential component of the electric field intensity is zero. Consider a gaussian cylinder of radius ρ and height h at the boundary, Applying Gauss law, & then integrating over the distinct surfaces we get. Flux experienced by the lateral surface is zero & Flux experienced by the bottom surface is zero as charge inside the conductor is zero. Therefore or. At the conductor dielectric interface normal component of the electric flux density is equal to the surface charge density. Dept of ECE, SJBIT 9

30 Problem Dept of ECE, SJBIT 30

31 Boundary condition for perfect dielectric: Consider a closed path abcda at the dielectric dielectric interface & h 0. The work done in moving a unit charge over a closed path is zero. Therefore, We know that the work done in moving a unit charge over a closed path is zero. Therefore,, and hence. The small contribution of the normal component of E due to h becomes negligible. Therefore,. & as D = E we get, or. At the dielectric dielectric boundary tangential component of the E is continuous where as tangential component of electric flux density is discontinuous. Consider a gaussian cylinder of radius ρ and height h at the boundary, Applying Gauss law, Dept of ECE, SJBIT 31

32 & then integrating over the distinct surfaces we get Flux experienced by the lateral surface is zero. Therefore. From which,.. For perfect dielectric, D N1 = D N, then E = 1 E 1. At the dielectric dielectric boundary normal component of the flux density is continuous. Normal components of D are continuous, The ratio of the tangential components,. And Or. The magnitude of D is given by,.. Dept of ECE, SJBIT 3

33 RECOMMENDED QUESTIONS 1. Define electric scalar potential. Establish the relationship between intensity and potential.. Discuss the boundary conditions between perfect dielectrics. 3. State & explain the principle of charge conservation. 4. Derive for energy stored in an electrostatic field. 5. Derive for energy expended in moving a point charge in an electric field. 6. Define Potential & potential difference. 7. Prove that E is Grad of V 8. Write a short note on dipole 9. Three point charges, 0.4 μc each, are located at (0,0,-1), (0,0,0) and (0,0,1) in free space. (a). Find an expression for the absolute potential as a function of Z along the lne x=0, y=1. (b) Sketch V(Z). Dept of ECE, SJBIT 33

34 UNIT 3 POISSONS AND LAPLACES EQUATION Syllabus: Derivation of Poisson s equation and Laplace s equation, Uniqueness theorem, Examples of the solutions Laplace Equations and Poisson s Equations REFERENCE BOOKS: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006.. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Laplace s & Poisson s equation: Laplace s & Poisson s equation enable us to find potential fields within regions bounded by known potentials or charge densities. Derivation of Laplace s & Poisson s equation: From Gauss law in point form, we have (1). By definition, D = E. & from gradient relationship, By substituting the above in equation 1, we get. Or (). For a homogeneous region in which is constant. Equation is poisson s equation. Dept of ECE, SJBIT 34

35 In rectangular co-ordinates, Therefore,.. If ρ v = 0, indicating zero volume charge density, but allowing point charges, line charges & surface charge density to exist at singular locations as sources of the field, then which is Laplace s equation. The operorator is called the Laplacian of V. In rectangular coordinates Laplace equation is, In cylindrical coordinates,, & in spherical coordinates, Every conductor produces a field for which V=0. In examples if it satisfies the boundary conditions and Laplace equation, then it is the only possible answer. Uniqueness theorem: Let us assume we have two solutions of Laplace equation, V 1 and V, both general functions of the coordinates used. Therefore V 1 = 0 and V = 0. From which. On the boundary, V b1 =V b. Let the difference between V 1 & V be V d. Therefore V d = V 1 -V. From Laplace equation, V d = V - V 1. On the boundary V d =0. Dept of ECE, SJBIT 35

36 From Divergence theorem,. Using vector identity, We get,. As V 1 = V,. Surface consists of boundaries and hence. Therefore. And. As. We obtain,. Example of solution of Laplace s equation: Example 1: For a Parallel plate capacitor: Let us assume V is a function of x. Laplace s equation reduces to,. Since V is not a function of y & z. Integrating the above equation twice we obtain, Where A & B are integration constants.. If V=0 at x=0 and V= V 0 at x = d, then, A= V 0 /d and B = 0. Dept of ECE, SJBIT 36

37 Therefore, Hence we have,. And the capacitance is Example : Capacitance of a co-axial cylindrical conductor: Assuming variation with respect to ρ Laplace equation becomes,. Integrating twice on both sides we obtain,.,. Assuming V = V 0 at ρ = A and V= 0 at ρ = B, We get. Dept of ECE, SJBIT 37

38 . Example 3: Spherical capacitor: Assuming variation with respect to r Laplace equation becomes, Integrating twice on both sides we obtain,., Assuming V = V 0 at θ = Π/ and V= 0 at θ = α, We get..,,, Dept of ECE, SJBIT 38

39 and. RECOMMENDED QUESTIONS 1. Derive Poisson s & Laplace s equation.. Using Laplace s equation, Prove that the potential distribution at any point in the region between two concentric cylinders of radii A & B as V=Voln ῤ/B /ln A/B 3. State and prove uniqueness theorem 4. Derive for Capacitance of Parallel plate capacitor 5. Derive for Capacitance of Concentric spherical capacitor. 6. Let V = xy z 3 and ε = ε 0. Given point P(1,,-1), Find (a) V at P; (b) E at P; (c) ρ v at P; (d) the equation of the equipotential surface passing through P; (e) the equation of the streamline passing through P; (f) Does V satisfy the Laplaces Equation Dept of ECE, SJBIT 39

40 UNIT 4 MAGNETOSTATIC FIELDS Syllabus: Biot- savart law, Amphere s circuitary law, curl, Stoke s theorem, Magnetic flux and flux density, Scalar and Vector magnetic potentials. REFERENCE BOOKS: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006.. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Introduction: Static electric fields are characterized by E or D. Static magnetic fields, are characterized by H or B. There are similarities and dissimilarities between electric and magnetic fields. As E and D are related according to D = E for linear material space, H and B are related according to B = H. A definite link between electric and magnetic fields was established by Oersted in 180. An electrostatic field is produced by static or stationary charges. If the charges are moving with constant velocity, a static magnetic (or magnetostatic) field is produced. A magnetostatic field is produced by a constant current flow (or direct current). This current flow may be due to magnetization currents as in permanent magnets, electron-beam currents as in vacuum tubes, or conduction currents as in current-carrying wires. The development of the motors, transformers, microphones, compasses, telephone bell ringers, television focusing controls, advertising displays, magnetically levitated high speed vehicles, Dept of ECE, SJBIT 40

41 memory stores, magnetic separators, and so on, involve magnetic phenomena and play an important role in our everyday life. There are two major laws governing magnetostatic fields: (1) Biot-Savart's law, and () Ampere's circuit law. Like Coulomb's law, Biot-Savart's law is the general law of magnetostatics. Just as Gauss's law is a special case of Coulomb's law, Ampere's law is a special case of Biot-Savart's law and is easily applied in problems involving symmetrical current distribution. BIOT SAVART's LAW: Biot-Savart's law states that the magnetic field intensity dh produced at a point P, as shown in Figure 1.1, by the differential current element I dl is proportional to the product I dl and the sine of the angle between the element and the line joining P to the element and is inversely proportional to the square of the distance R between P and the element. That is, I dl sin dh (1.1) R or KI dl sin dh (1.) R where, k is the constant of proportionality. In SI units, k = 1/4. So, eq. (1.) becomes dh I dl sin (1.3) 4R From the definition of cross product equation A x B = AB Sin AB a n, it is easy to notice that eq. (1.3) is better put in vector form as Idl ar Idl R dh (1.4) 3 4R 4R Dept of ECE, SJBIT 41

42 where R in the denominator is R and a R = (vector R/ R }. Thus, the direction of dh can be determined by the right-hand rule with the right-hand thumb pointing in the direction of the current, the right-hand fingers encircling the wire in the direction of dh as shown in Figure 1.(a). Alternatively, one can use the right-handed screw rule to determine the direction of dh: with the screw placed along the wire and pointed in the direction of current flow, the direction of advance of the screw is the direction of dh as in Figure 1.(b). Figure 1.1: Magnetic field dh at P due to current element I dl. Figure 1.: Determining the direction of dh using (a) the right-hand rule, or (b) the right-handed screw rule. It is customary to represent the direction of the magnetic field intensity H (or current I) by a small circle with a dot or cross sign depending on whether H (or I) is out of, or into, the page as illustrated in Figure 1.3. Dept of ECE, SJBIT 4

43 As like different charge configurations, one can have different current distributions: line current, surface current and volume current as shown in Figure 1.4. If we define K as the surface current density (in amperes/meter) and J as the volume current density (in amperes/meter square), the source elements are related as I dl K ds J dv (1.5) Thus, in terms of the distributed current sources, Biot-Savart law as in eq. (1.4) becomes H L Idl a 4R R (Line current) (1.6) H S KdS a 4R R (Surface current) (1.7) H V Jdv a 4R R (Volume current) (1.8) As an example, let us apply eq. (1.6) to determine the field due to a straight current carrying filamentary conductor of finite length AB as in Figure 1.5. We assume that the conductor is along the z-axis with its upper and lower ends respectively subtending angles Figure 1.3: Conventional representation of H (or I) (a) out of the page and (b) into the page. Dept of ECE, SJBIT 43

44 Figure 1.4: Current distributions: (a) line current (b) surface current (c) volume current. and 1 at P, the point at which H is to be determined. Particular note should be taken of this assumption, as the formula to be derived will have to be applied accordingly. If we consider the contribution dh at P due to an element dl at (0, 0, z), dh Idl R (1.9) 3 4R But dl = dz a z and R = a - za z, so Hence, dl x R = dz a (1.10) H I dz 4 z 3 a (1.11) Figure 1.5: Field at point P due to a straight filamentary conductor. Dept of ECE, SJBIT 44

45 Letting z = cot, dz = - cosec d, equation (1.11) becomes H I 4 1 cos ec d a 3 3 cos ec I a d sin 4 1 Or I H cos cos 1a (1.1) 4 The equation (1.1) is generally applicable for any straight filamentary conductor of finite length. Note from eq. (1.1) that H is always along the unit vector a (i.e., along concentric circular paths) irrespective of the length of the wire or the point of interest P. As a special case, when the conductor is semi-infinite (with respect to P), so that point A is now at O(0, 0, 0) while B is at (0, 0, ); 1 = 90, = 0, and eq. (1.1) becomes I H a (1.13) 4 Another special case is when the conductor is infinite in length. For this case, point A is at (0, 0, - ) while B is at (0, 0, ); 1 = 180, = 0. So, eq. (1.1) reduces to I H a (1.14) To find unit vector a in equations (1.1) to (1.14) is not always easy. A simple approach is to determine a from a a (1.15) a Dept of ECE, SJBIT 45

46 where a l is a unit vector along the line current and a is a unit vector along the perpendicular line from the line current to the field point. Illustration: The conducting triangular loop in Figure 1.6(a) carries a current of 10 A. Find H at (0, 0, 5) due to side 1 of the loop. Solution: This example illustrates how eq. (1.1) is applied to any straight, thin, current-carrying conductor. The key point to be kept in mind in applying eq. (1.1) is figuring out 1,, and a. To find H at (0, 0, 5) due to side 1 of the loop in Figure 1.6(a), consider Figure Figure 1.6: (a) conducting triangular loop (b) side 1 of the loop. 1.6(b), where side 1 is treated as a straight conductor. Notice that we join the Point of interest (0, 0, 5) to the beginning and end of the line current. Observe that 1, and are assigned in the same manner as in Figure 1.5 on which eq. (1.1) is based. Dept of ECE, SJBIT 46

47 cos 1 = cos 90 = 0, cos, = 5 9 To determine a is often the hardest part of applying eq. (1.1). According to eq. (1.15), a l = a x and a = a z, so Hence, a = a x x a z = -a y H (5) cos cos a 0 ( a ) 1 1 y = a y ma/m 9 Ampere's circuit law Ampere's circuit law states that the line integral of the tangential components of H around a closed path is the same as the net current I enc enclosed by the path In other words, the circulation of H equals I enc ; that is, H dl I (1.16) enc Ampere's law is similar to Gauss's law and it is easily applied to determine H when the current distribution is symmetrical. It should be noted that eq. (1.16) always holds whether the current distribution is symmetrical or not but we can only use the equation to determine H when symmetrical current distribution exists. Ampere's law is a special case of Biot-Savart's law; the former may be derived from the latter. By applying Stoke's theorem to the left-hand side of eq. (1.16), we obtain But I H dl ( H) ds (1.17) enc L S I enc J ds (1.18) S Dept of ECE, SJBIT 47

48 Comparing the surface integrals in eqs. (1.17) and (1.18) clearly reveals that x H = J (1.19) This is the third Maxwell's equation to be derived; it is essentially Ampere's law in differential (or point) form whereas eq. (1.16) is the integral form. From eq. (1.19), we should observe that X H = J 0; that is, magneto-static field is not conservative. APPLICATIONS OF AMPERE'S LAW Infinite Line Current Consider an infinitely long filamentary current I along the z-axis as in Figure To determine H at an observation point P, we allow a closed path pass through P. This path on, which Ampere's law is to be applied, is known as an Amperian path (analogous to the term Gaussian surface). We choose a concentric circle as the Amperian path in view of eq. (1.14), which shows that H is constant provided p is constant. Since this path encloses the whole current I, according to Ampere's law Figure 1.7: Ampere's law applied to an infinite filamentary, line current. H 1 a (1.0) As expected from eq. (1.14). Dept of ECE, SJBIT 48

49 MAGNETIC FLUX DENSITY The magnetic flux density B is similar to the electric flux density D. As D = 0 E in free space, the magnetic flux density B is related to the magnetic field intensity H according to B = 0 H (1.1) where, 0 is a constant known as the permeability of free space. The constant is in henrys/meter (H/m) and has the value of 0 = 4 x 10-7 H/m (1.) The precise definition of the magnetic field B, in terms of the magnetic force, can be discussed later. Figure 1.8: Magnetic flux lines due to a straight wire with current coming out of the page The magnetic flux through a surface S is given by B ds (1.3) S Where the magnetic flux is in webers (Wb) and the magnetic flux density is a webers/square meter (Wb/m ) or teslas. An isolated magnetic charge does not exit. Total flux through a closed surface in a magnetic field must be zero; that is, B ds 0 (1.4) Dept of ECE, SJBIT 49

50 This equation is referred to as the law of conservation of magnetic flux or Gauss' s law for magnetostatic fields just as D. ds = Q is Gauss's law for electrostatic fields. Although the magnetostatic field is not conservative, magnetic flux is conserved. By applying the divergence theorem to eq. (1.4), we obtain Or S B ds v B dv 0. B = 0 (1.5) This equation is the fourth Maxwell's equation to be derived. Equation (1.4) or (1.5) shows that magnetostatic fields have no sources or sinks. Equation (1.5) suggests that magnetic field lines are always continuous. TABLE 1.: Maxwell's Equations for Static EM Fields Differential (or Integral Form Point) Form. D = v ds S D v dv Gauss's law v Remarks. B = 0 B ds 0 Nonexistence of magnetic monopole S x E = 0 E dl 0 Conservativeness of electrostatic field x H = J H dl J L L s ds Ampere's law The Table 1. gives the information related to Maxwell's Equations for Static Electromagnetic Fields. MAGNETIC SCALAR AND VECTOR POTENTIALS We recall that some electrostatic field problems were simplified by relating the electric Potential V to the electric field intensity E (E = -V). Similarly, we can define a potential Dept of ECE, SJBIT 50

51 associated with magnetostatic field B. In fact, the magnetic potential could be scalar V m vector A. To define V m and A involves two important identities: x (V) = 0 (1.6). ( x A) = 0 (1.7) which must always hold for any scalar field V and vector field A. Just as E = -V, we define the magnetic scalar potential V m (in amperes) as related to H according to H = - V m if J = 0 (1.8) The condition attached to this equation is important and will be explained. Combining eq. (1.8) and eq. (1.19) gives J = x H = - x (- V m ) = 0 (1.9) since V m, must satisfy the condition in eq. (1.6). Thus the magnetic scalar potential V m is only defined in a region where J = 0 as in eq. (1.8). We should also note that V m satisfies Laplace's equation just as V does for electrostatic fields; hence, V m = 0, (J = 0) (1.30) We know that for a magnetostatic field, x B = 0 as stated in eq. (1.5). In order to satisfy eqs. (1.5) and (1.7) simultaneously, we can define the vector magnetic potential A (in Wb/m) such that Just as we defined B = x A (1.31) V dq 4 0 r (1.3) We can define A L I 0 dl 4R for line current (1.33) Dept of ECE, SJBIT 51

52 A S K 0 4R ds for surface current (1.34) A v J 0 4R dv for volume current (1.35) Illustration 1: Given the magnetic vector potential A = - /4 a z Wb/m, calculate the total magnetic flux crossing the surface = /, 1 m, 0 z 5m. Solution: Az B A a a, ds d dz a B ds 1 5 z0 1 1 d dz 4 (5) 15 4 = 3.75 Wb Illustration : Identify the configuration in figure 1.9 that is not a correct representation of I and H. Solution: Figure 1.9: Different I and H representations (related to Illustration ) Dept of ECE, SJBIT 5

53 Figure 1.9 (c) is not a correct representation. The direction of H field should have been outwards for the given I direction. RECOMMENDED QUESTIONS 1. State and prove Biot Savart s law.. Derive for H at a point due to an infinite long conductor. 3. Derive for H at a point due to finite straight conductor. 4. Derive for H on the axis of a circular loop. 5. State and prove Ampere s circuit law. 6. Mention the applications of Ampere s circuit law and derive any one of them in brief. 7. Define curl and discuss the significance of curl. 8. Derive for Ampere s law in point form. 9. Define magnetic flux & flux density. 10. Write a short note on magnetic scalar and vector potential. Dept of ECE, SJBIT 53

54 UNIT 5 MAGNETIC FORCES, MATERIALS AND INDUCTANCE Syllabus: (a).force on a moving charge and differential current element, Force between differential current elements, Force and Torque on a closed circuit, (b).magnetic materials and inductance, Magnetization and permeability, Magnetic boundary conditions, Magnetic circuits, Potential energy and forces on magnetic materials, Inductance and mutual inductance REFERENCE BOOKS: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7 th edition,006.. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, Electromagnetic Waves And Radiating Systems, Edward C. Jordan and Keith G Balmain, Prentice Hall of India / Pearson Education, nd edition, 1968.Reprint Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, nd edition, , Indian Reprint 001. Force on a Charged Particle According to earlier information, the electric force F e, on a stationary or moving electric charge Q in an electric field is given by Coulornb's experimental law and is related to the electric field intensity E as F e = QE (.1) This shows that if Q is Positive, F e and E have the same direction. A magnetic field can exert force only on a moving charge. From experiments, it is found that the magnetic force F m experienced by a charge Q moving with a velocity u in a magnetic field B is F m = Qu x B (.) This clearly shows that F m is perpendicular to both u and B. Dept of ECE, SJBIT 54

55 From eqs. (.1) and (.), a comparison between the electric force F e and the magnetic force F m can be made. F e is independent of the velocity of the charge and can perform work on the charge and change its kinetic energy. Unlike F e, F m depends on the charge velocity and is normal to it. F m cannot perform work because it is at right angles to the direction of motion, of the charge (F m.dl = 0); it does not cause an increase in kinetic energy of the charge. The magnitude of F m is generally small compared to F e except at high velocities. For a moving charge Q in the Presence of both electric and magnetic fields, the total force on the charge is given by or F = F e + F m F = Q (E + u x B) (.3) This is known as the Lorentz force equation. It relates mechanical force to electrical force. If the mass of the charged Particle moving in E and B fields is m, by Newton's second law of motion. du F m QE u B (.4) dt The solution to this equation is important in determining the motion of charged particles in E and B fields. We should bear in mind that in such fields, energy transfer can be only by means of the electric field. A summary on the force exerted on a charged particle is given in table.1. TABLE.1: Force on a Charged Particle State of Particle E Field B Field Combined E and B Fields Stationary QE - QE Moving QE Qu x B Q(E + u x B) The magnetic field B is defined as the force per unit current element Alternatively, B may be defined from eq. (.) as the vector which satisfies F m / q = u x B just as we defined electric field E as the force per unit charge, F e / q. Dept of ECE, SJBIT 55

56 Force between Two Current Elements Let us now consider the force between two elements I 1 dl 1 and I dl. According to Biot-Savart's law, both current elements produce magnetic fields. So we may find the force d(df 1 ) on element I 1 dl 1 due to the field db produced by element I dl as shown in Figure.1. As per equation df = I dl x B d(df 1 ) = I 1 dl 1 x db (.5) But from Biot-Savart's law, Hence, db 0I dl ar 1 (.6) 4R 1 1 0I1dl1 I dl ar 1 d( df1 ) (.7) 4R Figure.1: Force between two current loops. This equation is essentially the law of force between two current elements and is analogous to Coulomb's law, which expresses the force between two stationary charges. From eq. (.7), we obtain the total force F 1 on current loop 1 due to current loop shown Figure.1 as 0I1I F1 dl 1 4 L1 L R1 dl a R1 (.8) Dept of ECE, SJBIT 56

57 Although this equation appears complicated, we should remember that it is based on eq. (.5). It is eq. (8. 10) that is of fundamental importance. The force F on loop due to the magnetic field B 1 from loop 1 is obtained from eq. (.8) by interchanging subscripts 1 and. It can be shown that F = - F 1 ; thus F 1 and F obey Newton's third law that action and reaction are equal and opposite. It is worthwhile to mention that eq. (.8) was experimentally established by Qersted and Ampete; Biot and Savart (Ampere's colleagues) actually based their law on it. MAGNETIC TORQUE AND MOMENT Now that we have considered the force on a current loop in a magnetic field, we can determine the torque on it. The concept of a current loop experiencing a torque in a magnetic field is of paramount importance in understanding the behavior of orbiting charged particles, d.c. motors, and generators. If the loop is placed parallel to a magnetic field, it experiences a force that tends to rotate it. The torque T (or mechanical moment of force) on the loop is the, vector product of the force F and the moment arm r. That is, T = r x F (.9) and its units are Newton-meters. Let us apply this to a rectangular loop of length l and width w placed in a uniform magnetic field B as shown in Figure 8.5(a). From this figure, we notice that dl is parallel to B along sides 1 and 34 of the loop and no force is exerted on those sides. Thus F I 3 dl B I dl 1 4 B l 0 I dz az B I dz az l B Dept of ECE, SJBIT 57

58 Figure.: Rectangular planar loop in a uniform magnetic field. or F= F 0 F 0 = 0 (.10) Where, F 0 = I Bl because B is uniform. Thus, no force is exerted on the loop as a whole. However, F 0 and F 0 act at different points on the loop, thereby creating a couple. If the normal to the plane of the loop makes an angle with B, as shown in the cross-sectional view of Figure.(b), the torque on the loop is T = F 0 w sin or T = B I l w sin (.11) But lw = S, the area of the loop. Hence, T = BIS sin (.1) We define the quantity m = I S a n (.13) as the magnetic dipole moment (in A/M ) of the loop. In eq. (.13), a n is a unit normal vector to the plane of the loop and its direction is determined by the right-hand rule: fingers in the direction of current Hand thumb along a n. The magnetic dipole moment is the product of current and area of the loop; its reaction is normal to the loop. Introducing eq. (.13) in eq. (.1), we obtain Dept of ECE, SJBIT 58

59 T = m x B (.14) Stoke's theorem Stoke's Theorem relates a line integral to the surface integral and vice-versa, that is H dl H) C S ( ds (3.1) Force on a moving charge due to electric and magnetic fields If there is a charge or a moving charge, Q in an electric field, E, there exists a force on the charge. This force is given by F E = QE (3.) If a charge, Q moving with a velocity, V is placed in a magnetic field, B (=H), then there exists a force on the charge (Fig. 3.1). This force is given by F H = Q(V x B) (3.3) B = magnetic flux density, (wb/m ) V = velocity of the charge, m/s Fig. 3.1: Direction of field, velocity and force If the charge, Q is placed in both electric and magnetic fields, then the force on the charge is F =Q (E + V x B) (3.4) This equation is known as Lorentz force equation. Dept of ECE, SJBIT 59